What rotation is required to remove the #xy#-term from the conic #x^2+2sqrt(3)xy+3y^2+2sqrt(3)x-2y=0#?

1 Answer
Aug 9, 2017

A rotation through #(2pi)/6# radians, (or #60^o#) will remove the #xy#-term

The new translated equation is:

# X^2 - Y=0 #

Which represents a paraobla

Explanation:

We have:

# x^2+2sqrt(3)xy+3y^2+2sqrt(3)x-2y=0 #

Comparing to the conic equation in standard form:

# Ax^2 + Bxy + Cy^2 + Dx +Ey + F = 0 #

We have:

# A=1; \ B=2sqrt(3); \ C=3; \ D=2sqrt(3); \ E=-2; \ F=0 #

The rotation angle, #theta#, that we seek to translate from the #xy#-plane to the #XY#-plane (say) and remove the #xy#-term is given by:

# cot 2theta = (A-C)/B = (1-3)/(2sqrt(3) = -1/(sqrt(3) #
# :. 2theta = (2pi)/3 => theta = (2pi)/6 #

We will need #sin theta#, and #cos theta#, so let us calculate those also:

# sin theta = (sqrt(3))/2; \ \ \ \ cos theta = 1/2 #

The rotation matrix to transform from #xy#-coordinate axis into the #XY#-coordinate axis is therefore:

# ( (x),(y) ) = ( (cos theta, -sin theta), (sin theta, cos theta) ) ( (X),(Y) )#
# " " = ( (1/2, -(sqrt(3))/2), ((sqrt(3))/2, 1/2) ) ( (x'),(y') )#

Leading to the transformation equations:

# x = 1/2 X-sqrt(3)/2 Y; \ \ \ \ y = sqrt(3)/2 X + 1/2 Y #

We can substitute these transformation equations into the original equation to get the new equation (which should be deficient in an #XY#-term)

# (1/2 X-sqrt(3)/2 Y)^2+2sqrt(3)(1/2 X-sqrt(3)/2 Y)(sqrt(3)/2 X + 1/2 Y)+3(sqrt(3)/2 X + 1/2 Y)^2+2sqrt(3)(1/2 X-sqrt(3)/2 Y)-2(sqrt(3)/2 X + 1/2 Y)=0 #

We can remove the fractions to make the algebra slightly less tedious, by multiplying by #4#

# (X-sqrt(3)Y)^2 + 2sqrt(3)(X-sqrt(3)Y)(sqrt(3)X + Y) + 3(sqrt(3) X + Y)^2 + 4sqrt(3)(X-sqrt(3)Y) - 4(sqrt(3) X + Y) = 0 #

And now we multiply out and collect terms:

# (X^2-2sqrt(3)XY+3Y^2) + 2sqrt(3)(sqrt(3)X^2+XY-3XY-sqrt(3)Y^2) + 3(3X^2+2sqrt(3)XY+Y^2) + 4sqrt(3)(X-sqrt(3)Y) - 4(sqrt(3) X + Y)=0 #

# :. X^2-2sqrt(3)XY+3Y^2 + 6X^2-4sqrt(3)XY-6Y^2 + 9X^2 + 6sqrt(3)XY + 3Y^2 + 4sqrt(3)X - 12Y - 4sqrt(3)X - 4Y=0 #

# :. 16X^2 - 16Y=0 #

# :. X^2 - Y=0 #

Which in the #XY#-plane we can easily identify as a parabola.

The graphs of the two equation (on the same image) as as follows:

enter image source here

And we can graphically confirm that the original equation is the same parabola rotated through #(2pi)/6# radians, (or #60^o#)